The Disk or Shell Method Calculus

Question

Consider the region bounded by the x-axis and the function f(x)=-x² +3x -2. Revolve the region about the y-axis and find the volume.

Answer 1

METHOD OF CYLINDRICAL SHELLS: The graph of y = -x² + 3x − 2 shows intersection with the x-axis at (1,0) and (2,0). Using the Cylindrical Shell Method, write Volume or V = 2π∫xf(x)dx or, in this problem, V = 2π∫_{(from x = 1 to x = 2)}x(-x² + 3x − 2)dx.

Rewrite as 2π∫_{(from x = 1 to x = 2)}(-x³ + 3x² − 2x)dx or 2π[-x⁴/4 + x³ − x²|_{(from x = 1 to x = 2)}].

Evaluation from x = 1 to x = 2 yields 2π[-2⁴/4 + 2³ − 2² − (-1⁴/4 + 1³ − 1²)] or 2π[-4 + 8 − 4 + 1/4 −1 + 1] which simplifies to 2π[1/4] or π/2 cubic units.

METHOD OF DISKS OR CIRCULAR WASHERS: Using the Disk Method along the y-axis, write y = -x² + 3x − 2 or -x² + 3x − (2 + y) = 0.

Using the Quadratic Formula, obtain x = [-3 ± √(3² − 4(-1)(-2 − y))]/-2 which reduces to x = 1.5 ± [√(9 − 8 − 4y)]/-2 or x = 1.5 ±(1/2)√(1 − 4y).

{Note that, for y = 0, x = 1.5 ± 1/2 which gives x = 1 or x = 2; these are the limits of integration for the Cylindrical Shell Method shown above.}

Examination of the graph of y = -x² + 3x − 2 shows that y = 0 everywhere on the x-axis and y = 0.25 at the function's maximum (at the point (1.5,0.25)). The limits of integration for the Disk Method will then be y = 0 and y = 0.25.

Next write V = π∫_{(from y = 0 to y = 0.25)}[(1.5 + (1/2)(1 − 4y)^(1/2))² − (1.5 − (1/2)(1 − 4y)^(1/2))²]dy. This simplifies to V = π∫_{(from y = 0 to y = 0.25)}[2(1.5)(1 − 4y)^(1/2)]dy.

Integrate this last expression to obtain V = π[(3/-6)(1 − 4y)^(3/2)|_{(from y = 0 to y = 0.25)}] or π[0 − (3/-6)], which gives π/2 cubic units in agreement with the result given by the Cylindrical Shell Method shown above.

Answer 2

So, if you sketch the parabola, y = -x² + 3x - 2, you'll have a parabola opening downwards hitting the x - axis at x = 1, 2. If you draw your rectangles vertically, and spin those rectangles around the y - axis, you'll have cross sections that are cylinders (or shells). Therefore your cross sectional area is 2π(r)(h), where in this case, r is the distance from your rectangular strip (which is at a random point x) and the y - axis (x = 0) Thus, your radius r = x. The height of your cylinder is just the height of your parabola, i.e. h = f(x). Putting this all together, you have the following:

V =2π∫_a^brh dx

= 2π ∫₁²x(-x² + 3x - 2) dx

= 2π ∫₁^2∈(-x³ + 3x² - 2x) dx

= 2π (-1/4 x⁴ + x³ - x²)Ι₁²

= 2π [(-4 + 8 - 4) - (-1/4 + 1 - 1)]

= 2π (1/4)

= π / 2

Hope this helps.

Mr. K